Question: Solve for $x$ and $y$ using substitution. ${3x+4y = 10}$ ${x = 3y-1}$
Solution: Since $x$ has already been solved for, substitute $3y-1$ for $x$ in the first equation. ${3}{(3y-1)}{+ 4y = 10}$ Simplify and solve for $y$ $9y-3 + 4y = 10$ $13y-3 = 10$ $13y-3{+3} = 10{+3}$ $13y = 13$ $\dfrac{13y}{{13}} = \dfrac{13}{{13}}$ ${y = 1}$ Now that you know ${y = 1}$ , plug it back into $\thinspace {x = 3y-1}\thinspace$ to find $x$ ${x = 3}{(1)}{ - 1}$ $x = 3 - 1$ ${x = 2}$ You can also plug ${y = 1}$ into $\thinspace {3x+4y = 10}\thinspace$ and get the same answer for $x$ : ${3x + 4}{(1)}{= 10}$ ${x = 2}$